Question: Divide the following complex numbers. $ \dfrac{-16-12i}{-4-3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4+3i}$ $ \dfrac{-16-12i}{-4-3i} = \dfrac{-16-12i}{-4-3i} \cdot \dfrac{{-4+3i}}{{-4+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-16-12i) \cdot (-4+3i)} {(-4-3i) \cdot (-4+3i)} = \dfrac{(-16-12i) \cdot (-4+3i)} {(-4)^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-16-12i) \cdot (-4+3i)} {(-4)^2 - (-3i)^2} = $ $ \dfrac{(-16-12i) \cdot (-4+3i)} {16 + 9} = $ $ \dfrac{(-16-12i) \cdot (-4+3i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-16-12i}) \cdot ({-4+3i})} {25} = $ $ \dfrac{{-16} \cdot {(-4)} + {-12} \cdot {(-4) i} + {-16} \cdot {3 i} + {-12} \cdot {3 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{64 + 48i - 48i - 36 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{64 + 48i - 48i + 36} {25} = \dfrac{100 + 0i} {25} = 4 $